Find the ratio of the frequency of the damped oscillator

[kreil]

Homework Statement

Consider a damped harmonic oscillator. After four cycles the amplitude of the oscillator has dropped to 1/e of its initial value. Find the ratio of the frequency of the damped oscillator to its natural frequency.

Homework Equations

$$T=\frac{ 2\pi}{w_1}$$

$$w_1^2=w_0^2- {\Beta}^2$$

The Attempt at a Solution

I know the answer is $$\frac{w_1}{w_0}=\frac{8 {\pi}}{\sqrt{64 {\pi}^2+1}}$$

However, I think I am missing an equation here because using only the two above i cannot obtain this answer. If someone could point me in the right direction I would appreciate it!

Josh

 

[cristo]

Well, firstly you should explain your symbols. I presume you are using w to mean omega; the frequency of the oscillator. What is the definition of the natural frequency?

 

[m2003]

:

I would first clarify any uncertainties or missing information in the given problem. Is the damping factor, Beta, known? Is the initial amplitude of the oscillator given? Without this information, it is difficult to verify the given answer or provide a complete solution.

Assuming that Beta and the initial amplitude are known, the ratio of the frequency of the damped oscillator to its natural frequency can be found using the equation:

$\frac{w_1}{w_0}=\sqrt{1-\frac{1}{e^2}}=\sqrt{1-\frac{1}{e^{2{\Beta}T}}}=\sqrt{1-\frac{1}{e^{2{\Beta}\frac{4\pi}{w_0}}}}$

Where T is the time for four cycles, which is equal to $\frac{4\pi}{w_0}$.

This equation can be derived from the equation for the amplitude of a damped harmonic oscillator, which is given by:

$A(t)=A_0e^{-{\Beta}t}$

Setting t equal to the time for four cycles, we get:

$\frac{A(t)}{A_0}=\frac{1}{e^{\Beta T}}=\frac{1}{e^{\Beta \frac{4\pi}{w_0}}}$

Solving for $\frac{w_1}{w_0}$, we get the same answer as the one given in the problem. Therefore, the given ratio is correct, but more information is needed to provide a complete solution.